\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)

\(C=\frac{1}{100}-1+\frac{1}{100}\)

\(C=\frac{-49}{50}\)

C = 1/100 - 1/100.99 - 1/99.98 - 1/98.97 - ... - 1/3.2 - 1/2.1

C = 1/100 - (1/100.99 + 1/99.98 + 1/98.97 + ... + 1/3.2 + 1/2.1)

C = 1/100 - (1/1.2 + 1/2.3 + ... + 1/98.99 + 1/99.100)

C = 1/100 - (1 - 1/2 + 1/2 - 1/3 + ... + 1/98 - 1/99 + 1/99 - 1/100)

C = 1/100 - (1 - 1/100)

C = 1/100 - 99/100

C = -98/100 = -49/50

\(c=\frac{1}{100}-\frac{1}{100.98}\frac{1}{99.98}\frac{1}{98.97}-......-\frac{1}{3.2}-\frac{1}{2.1}\)=\(\frac{1}{100}-\left[\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right]\) =\(\frac{1}{100}-\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{98}+\frac{1}{99}-\frac{1}{100}\right]\)=\(\frac{1}{100}-\left[1-\frac{1}{100}\right]=\frac{1}{100}-\frac{99}{100}=\frac{-98}{100}=\frac{49}{50}\)

\(C=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+\frac{1}{98.97}+..+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

Đặt biểu thức trong ngoặc đơn là B ta có

\(B=\frac{100-99}{100.99}+\frac{99-98}{99.98}+\frac{98-97}{98.97}+...+\frac{3-2}{3.2}+\frac{2-1}{2.1}\)

\(B=\frac{1}{99}-\frac{1}{100}+\frac{1}{98}-\frac{1}{99}+\frac{1}{97}-\frac{1}{98}+...+\frac{1}{2}-\frac{1}{3}+1-\frac{1}{2}\)

\(B=1-\frac{1}{100}=\frac{99}{100}\)

\(C=\frac{1}{100}-\frac{99}{100}=-\frac{98}{100}\)

C = \(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-.....-\frac{1}{2.1}\)

C = \(\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}\right)\)

C = \(\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

C = \(\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}\)

C = \(\frac{-49}{50}\)

-49/50

=>\(C=\frac{1}{100}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\right)\)

=>\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)

=>\(C=\frac{1}{100}-\frac{99}{100}=-\frac{98}{100}=-\frac{49}{50}\)

Vậy......

\(\frac{1}{\left(a+1\right)a}=\frac{1}{a}-\frac{1}{a+1}\)   
Áp dụng đẳng thức trên ta tính ĐƯỢC:
 A= 1/100-(1/99-1/100+1/98-1/99+...+1/2-1/3+1/1-1/2)
   =1/100-(-1/100+1)

   =1/50+1=51/50

\(A=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(A=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(A=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)

\(A=\frac{1}{100}-\frac{99}{100}\)

\(A=\frac{-98}{100}=-\frac{49}{50}\)

C = 1/100 - 1/100.99 - 1/99.98 - 1/98.97 - ... - 1/3.2 - 1/2.1

C = 1/100 - ( 1/100.99 + 1/99.98 + 1/98.97 + ... + 1/3.2 + 1/2.1)

C = 1/100 - ( 1/1.2 + 1/2.3 + ... + 1/97.98 + 1/98.99 + 1/99.100)

C = 1/100 - ( 1 - 1/2 + 1/2 - 1/3 + .... + 1/97 - 1/98 + 1/98 - 1/99 + 1/99 - 1/100)

C = 1/100 - ( 1 - 1/100)

C = 1/100 - 99/100

C = -98/100 = -49/50

\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}.\)

\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}=...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(=-\left(1-\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+\frac{1}{3}-...-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right).\)

\(=-\left(1-\frac{1}{100}\right)=-\frac{99}{100}\)

chúc bạn học tốt

Trả lời

1/100.99-1/99.98-1/98.97-...-1/3.2-1/2.1

=1/100-1/1

=1/100-100/100

=-99/100.

\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=\frac{1}{100.99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\right)\)

\(=\frac{1}{100.99}-\left(1-\frac{1}{99}\right)\)

\(=\frac{1}{100.99}-\frac{98}{99}\)

\(=\frac{1}{100.99}-\frac{9800}{99.100}\)

\(=\frac{-9799}{9900}\)

Ta có:\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=\frac{1}{9900}-\left(\frac{1}{99.98}+\frac{1}{98.97}+\frac{1}{97.96}+....+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

\(=\frac{1}{9900}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\right)\)

\(=\frac{1}{9900}-\left(1-\frac{1}{99}\right)\)

\(=\frac{1}{9900}-\frac{98}{99}=-\frac{9799}{9900}\)

\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=\frac{1}{100.99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}\right)\)

\(=\frac{1}{9900}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\right)\)

\(=\frac{1}{9900}-\left(1-\frac{1}{99}\right)\)

\(=\frac{1}{9900}-\frac{98}{99}=-\frac{9799}{9900}\)

\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=\frac{1}{100.99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}\right)\)

\(=\frac{1}{9900}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\right)\)

\(=\frac{1}{9900}-\left(1-\frac{1}{99}\right)\)

\(=\frac{1}{9900}-\frac{98}{99}=-\frac{9799}{9900}\)

1) Tính:

\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-....-\frac{1}{3.2}\)\(-\frac{1}{2.1}\)

\(=\frac{1}{100}-\frac{1}{99}-\frac{1}{99}-\frac{1}{98}-\frac{1}{98}-\frac{1}{97}-\)\(...-\frac{1}{3}-\frac{1}{2}-\frac{1}{2}-\frac{1}{1}\)

\(=\frac{1}{100}-\frac{1}{1}\)

\(=-\frac{99}{100}\)